Telling
Time by the Stars - Sidereal Time

Problem:Let
the vernal equinox occur at noon solar time on March 21 of a certain year.
Estimate the sidereal time at 3:00 pm solar time on November 29 of the
same year.

Solution:
Fix the earth-sun line in [inertial] space. Let the earth rotate
on its axis once a day, and let the celestial sphere rotate about the
celestial poles once a year. As viewed from the north celestial pole,
the earth will rotate counter clockwise, and the celestial sphere, clockwise.
The rotation rate of the earth is one rotation every 24 solar hours (length
of the solar day). The rotation rate of the celestial sphere is one rotation
every tropical year (365.2422 days). The relative rotation rate of the
earth and celestial sphere is

360o/day
+ 360o/(365.2422 days)

= 360o/day
+ 0.9856o/day

= 360.9856o/day

Thus,
relative to the earth, the celestial sphere completes one rotation (sidereal
day) in something less than a solar day. In fact, the sidereal day is
just

(360/360.9856)
x 24 hours = 23.9345 hours

= 23
hr 56 min 4.2 sec

i.e.,
approximately 3 min 56 sec shorter than the solar day. Another way to
show the same thing is to form the ratio of the length of the sidereal
day to that of the solar day:

Now we
may solve the problem of estimating the sidereal time on 3:00 pm, November
29. From noon on March 21 to noon on November 29 is 253 solar days. From
noon to 3:00 pm on November 29 is an additional 3/24 = 0.125 solar days.
Hence, the total elapsed time from the vernal equinox to 3:00 pm on november
29 is